Let $g(x)=6\text{sin}(x)-8e^x$. $g'(x)=$
Explanation: Recall that ${\dfrac{d}{dx}[e^x]=e^x}$ and ${\dfrac{d}{dx}[\text{sin}(x)]=\text{cos}(x)}$. $\begin{aligned} g'(x)&=\dfrac{d}{dx}[6\text{sin}(x)-8e^x] \\\\ &=6{\dfrac{d}{dx}[\text{sin}(x)]}-8{\dfrac{d}{dx}[e^x]} \\\\ &=6\cdot{\text{cos}(x)}-8{e^x} \end{aligned}$ In conclusion, $g'(x)=6\text{cos}(x)-8e^x$